In this video, we’re going to talk about how to find the determinant of a 4×4 matrix. Now the first thing you want to do is identify the row or the column with the greatest number of zeros. In this example problem, we can’t find any rows or columns with two or more zeros. So I’m going to start with the first row, just to keep things simple.

Now the four numbers that you see here in the first row, these are going to be used as the coefficients for when we form the 3×3 matrix. So let’s start with the first one. The first coefficient is 1. And it’s located in row 1, column 1. So we’re going to eliminate row 1 and column 1. And what’s left over are the numbers that we need for the 3×3 matrix that we’re about to write. And so that was 5, 0, 3, 2, 3, 5, 1, negative 2, 3.

Now let’s move on to the second number in the first row. Notice that we have a positive one here. The next one is going to be minus. The signs will alternate. So minus the next number, which is 0. And then we’re going to write the next 3×3 matrix. So 0 is in row 1, column 2. So the numbers that we have, 2, negative 1, 2, and then 0, 3, 3, 5, negative 2, 3, we’re going to put that in the next 3×3 matrix. So that was 2, 0, 3, negative 1, 3, 5, 2, negative 2, 3.

So just to review, the 0 was in row 1, column 1, and you can see the numbers here. 2, 0, 3, that’s in the second row. In the third row we have negative 1, 3, 5. And then in the last row we have 2, negative 2, 3.

Now let’s move on to the next number in row 1. So that’s going to be 4. And it’s going to be positive 4 because the previous coefficient was negative. So 4 is in row 1, column 3. And so that’s going to leave behind the numbers 2, 5, 3, negative 1, 2, 5, and then 2, 1, 3. I’m taking this one step at a time so you can follow along.

Now we’re going to move on to the last number in the first row. Now the sign is going to change from positive to negative. So it’s going to be negative, and then the number that we see here, minus 6. So negative 6 is in the first row, fourth column. So we’re left over with 2, 5, 0, negative 1, 2, 3, 2, 1, negative 2.

Now here’s the reason why you want to identify the row or the column with the greatest number of zeros. Because 0 times this determinant is going to be 0. And thus, this will simplify the work that you need to do. Because you need to find the determinant of each of these 3 by 3 matrices. So the more zeros that you have, the less work that you’ll have to do. Keep that in mind.

Now let’s evaluate the 3 by 3 matrices one at a time. So let’s start with the first one. 5, 0, 3, 2, 3, 5, 1, negative 2, 3. So let’s use the first row. So we have 5 as the first number. If we take out row 1, column 1, we’re left with a 2 by 2 matrix. So that’s going to be 3, 5, negative 2, 3. And then it’s going to be minus the next number, 0. And 0 is in row 1, column 2. It leaves behind 2, 5, 1, 3. And then plus the last number in the first row, which is 3. And that’s in row 1, column 3, which leaves behind 2, 3, 1, negative 2.

Now to evaluate a 2 by 2 matrix, what you need to do is you need to multiply the first diagonal going in that direction. 3 times 3 is 9. And then minus the second diagonal going in that direction. Negative 2 times 5 is negative 10. 0 times that determinant is 0. And then it’s going to be plus 3. 2 times negative 2 is negative 4. Minus 1 times 3, which is positive 3. So this is 9 plus 10, which is 19. And then negative 4 minus 3 is negative 7. 5 times 19 is 95. 3 times 7 is 21. 95 minus 21 is 74. So that’s the value, or the determinant rather, of this particular 3 by 3 matrix.

So now let’s repeat the process for the other two 3 by 3 matrices. So let’s move on to the next one. 2, 5, 3, negative 1, 2, 5, and then 2, 1, 3. So let’s use the first row again. We’ll start with the first number, 2. And it leaves behind 2, 5, 1, 3 in the 2 by 2 matrix. Then we need to alternate the sign. So we’re going to use the second number, but we’re going to put a negative sign in front of that. And we’re going to have the numbers negative 1, 5, 2, 3. Next, the last number in row 1 is 3. And if we eliminate row 1, column 3, we’re left over with negative 1, 2, 2, 1.

So we’re going to have 2 times 3, which is 6, minus 1 times 5, which is 5. And then it’s going to be minus 5. And then we have negative 1 times 3, that is negative 3, minus 2 times 5, which is positive 10. And the next, negative 1 times 1 is negative 1. And then minus 2 times 2, which is 4. 6 minus 5 is 1. Negative 3 minus 10 is negative 13. Negative 1 minus 4 is negative 5. Negative 5 times negative 13, that’s positive 65. 3 times negative 5 is negative 15. 65 minus 15 is 50. And 2 plus 50 is 52. So this is plus 4 times 52. Don’t forget the 4 here.

Now let’s put a zero for this particular 3×3 matrix. Now let’s work on the last one. So the numbers are 2, 5, 0, negative 1, 2, 3, 2, 1, negative 2. So starting with the first number, 2, if we eliminate the first row and the first column, we’re left with 2, 3, 1, negative 2. We’re left with 2, 3, 1, negative 2. And then minus 5, eliminating row 1, column 2, we’re left with negative 1, 3, 2, negative 2. And then plus 0, and we’re left with negative 1, 2, 2, 1.

So now if we multiply 2 by negative 2, that’s negative 4, minus 1 times 3, which is 3. And then minus 5. Next we have negative 1 times negative 2, which is 2, minus 2 times 3, and that’s 6. And of course this is 0, so we can get rid of that. Negative 4 minus 3 is negative 7. 2 minus 6 is negative 4. And then we have 2 times negative 7, which is negative 14. Negative 5 times negative 4, that’s 20